Question

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ?

Answer 1

Net acceleration is due to braking and centripetal acceleration

Due to Braking,

$a_T=0.5m/s^2$

Speed of the cyclist, $v=27km/h=7.5m/s$

Radius of the circular turn, $r=80m$
Centripetal acceleration is given as:

$a_c=\frac{V^2}{r}$

$=\left({7.5}^2\right)/80=0.70m/s^2$
Since the angle between $a_c$ and $a_T$ is 90⁰, the resultant acceleration a is given by:

$a=(a_c^2+a_T^2{)}^{1/2}$

$a=({0.7}^2+{0.5}^2{)}^{1/2}=0.86m/s^2$

$tan\theta =\frac{a_c}{a_T}$

where $\theta$ is the angle of the resultant with the direction of the velocity.

$tan\theta =\frac{0.7}{0.5}=1.4$

$\theta =ta{n}^{-1}\left(1.4\right)={54.56}^0$